A man driving a car leaves a point A drives up to 32.5 km in a direction of 070. A cyclist leaves the same point in a direction 130 travelling. After some few hours both drivers are 80km apart. Use this information to answer 3 questions.
(1). What is the distance covered by the cyclist at this time in 2d.p.
(2). Find the bearing of Cyclist from the Car. correct to 1dp.
(3). Find the shortest distance between the car and the line of path of the cyclist, in 2 dp.
this is just like the airplane problem. Study the solution I provided there, and if you get stuck, come on back with what you have tried.
I'm sorry but I didn't see the solution. Mr Oobleck
Who ever responds first and correctly gets $100
To answer these three questions, we can use vectors and trigonometry. Let's break down the problem step by step:
1. The distance covered by the cyclist can be found using the Pythagorean theorem. We have a right-angled triangle formed by the car, cyclist, and the distance between them. The distance between them is 80 km, and the car has already covered 32.5 km. Therefore, the distance covered by the cyclist is:
Distance_cyclist = √(80^2 - 32.5^2) km
Distance_cyclist ≈ √(6400 - 1056.25) km
Distance_cyclist ≈ √5343.75 km ≈ 73.08 km (approximated to 2 decimal places)
So, the distance covered by the cyclist is approximately 73.08 km.
2. To find the bearing of the cyclist from the car, we need to use trigonometry. We have an isosceles triangle formed by the car, cyclist, and their distance of 80 km. The angle between the car's direction and the line connecting the car and the cyclist is 130 - 70 = 60 degrees.
Using the sine rule, we can find the bearing of the cyclist:
sin(Bearing) / 80 = sin(60) / 73.08
Bearing ≈ arcsin((sin(60) / 73.08) * 80) ≈ 0.8281 radians
Converting radians to degrees:
Bearing ≈ 0.8281 * (180 / π) ≈ 47.42 degrees (rounded to 1 decimal place)
So, the bearing of the cyclist from the car is approximately 47.4 degrees.
3. The shortest distance between the car and the line of path of the cyclist can be found by drawing perpendicular lines from the car to the line connecting the car and the cyclist. This forms a right-angled triangle.
The shortest distance can be found using the formula for the distance between a point and a line. It is given by:
Distance_shortest = |(y2 - y1)x0 - (x2 - x1)y0 + x2y1 - x1y2| / √((x2 - x1)^2 + (y2 - y1)^2)
Where (x1, y1) and (x2, y2) are the coordinates of two points on the line (the car and the cyclist), and (x0, y0) are the coordinates of the car's position.
Plugging in the values:
(x1, y1) = (0, 0)
(x2, y2) = (32.5, 80)
(x0, y0) = (32.5, 0)
Distance_shortest = |(80)x32.5 - (32.5)x0 + 32.5x0 - 0x80| / √((32.5 - 0)^2 + (80 - 0)^2)
Distance_shortest = |2600 - 0 + 0 - 0| / √(1056.25 + 6400)
Distance_shortest = 2600 / √(7456.25)
Distance_shortest ≈ 2600 / 86.32 ≈ 30.11 km (approximated to 2 decimal places)
So, the shortest distance between the car and the line of path of the cyclist is approximately 30.11 km.